Show this function is associative (or provide a counter example)

[Dick]

Well, it does give $\displaystyle \ f(x,y)=\sinh(\text{arcsinh}(x)+\text{arcsinh}(y)) \$ rather easily.

To get texy inverse hyperbolics I use

\text{arcsinh} .

Yeah, I see what you are doing now. And thanks for the texy tip.

 

[ArcanaNoir]

Just confirmed with the professor, hyperbolic trig is the way to go. *still trying to get it*

 

[I like Serena]

Just confirmed with the professor, hyperbolic trig is the way to go. *still trying to get it*

Use $\cosh^2 a - \sinh^2 a = 1$.

Btw, according to wiki, $\text{arcsinh}$ is a misnomer. It should be $\text{arsinh}$.

 

[ArcanaNoir]

Okay, you guys have been wonderful, especially Dick and Micro. Although I'm beginning to suspect Micro is a genius, in which case he sneezes and answers pop out of him, so it's not like he has to try really hard. :P j/k I appreciate your tireless efforts Micro! (not kidding about thinking you're a genius...)
Anyway I'm trying to use Micro's hint but I feel like I'm slipping past the part where I'm supposed to rearrange the expressions to show they are equivalent. I will show my simplification for the expressions, maybe someone can point out where I was supposed to do something trig-y.

I have verified that \cosh (\sinh ^{-1}(x))=\sqrt{1+x^2} and that f(x,y)=\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))

A:
(x\ast y)\ast z = \sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))\ast z \\ <br /> = \sinh (\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]+\sinh ^{-1} (z)) \\<br /> =\sinh [\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]]\cosh (\sinh ^{-1} (z))+\\<br /> \cosh [\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]]\sinh (\sinh ^{-1} (z)) \\ <br /> =\sinh [\sinh ^{-1} [\sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1} (y))+\cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1} (y))]]\cdot \sqrt{1+z^2} +\\<br /> \cosh [\sinh ^{-1} [ \sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1}(y))+\cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1} (y))]]\cdot z \\<br /> \sinh [\sinh ^{-1} [x\sqrt{1+y^2}+y\sqrt{1+x^2}]]\sqrt{1+z^2}+\cosh [\sinh ^{-1} [x\sqrt{1+y^2}+y\sqrt{1+x^2}]]z

B:
x\ast (y\ast z) = \sinh (\sinh ^{-1} (x)+\sinh ^{-1}(y\ast z)) \\<br /> =\sinh (\sinh ^{-1}(x)+\sinh ^{-1} [\sinh (\sinh ^{-1}(y) + \sinh ^{-1}(z))]) \\<br /> =\sinh [\sinh ^{-1} (x) + \sinh ^{-1} [\sinh (\sinh ^{-1}(y))\cosh (\sinh ^{-1} (z))+\\<br /> \cosh (\sinh ^{-1} (y))\sinh (\sinh ^{-1} (z))]] \\<br /> =\sinh [ \sinh ^{-1}(x)+\sinh ^{-1} [y\sqrt{1+z^2}+z\sqrt{1+y^2}]] \\<br /> =\sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1} [y\sqrt{1+z^2}+z\sqrt{1+y^2}])+\\<br /> \cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1}[y\sqrt{1+z^2}+z\sqrt{1+y^2}]) \\<br /> =x\sqrt{1+(y\sqrt{1+z^2}+z\sqrt{1+y^2})^2}+[y\sqrt{1+z^2}+z\sqrt{1+y^2}]\sqrt{1+x^2}

So it seems to me that this isn't going to end any better than when I didn't use hyp. trig, Hence why I think I'm missing the critical point.

 

[Dick]

You've already typed too much. Stop! Look at your second line. The first argument to the outer sinh is arcsinh(sinh('something')). Just replace that with 'something'. You are overshooting!

 

[ArcanaNoir]

You've already typed too much. Stop! Look at your second line. The first argument to the outer sinh is arcsinh(sinh('something')). Just replace that with 'something'. You are overshooting!

whoops! lol didnt see you down here, was just trying to fix my tex :) *looking at second line*

Umm maybe I don't see what you mean, I can't simplify directly when there is a sum involved, right?

 

[I like Serena]

B:
x\ast (y\ast z) = \sinh (\sinh ^{-1} (x)+\sinh ^{-1}(y\ast z)) \\<br /> =\sinh (\sinh ^{-1}(x)+\sinh ^{-1} [\sinh (\sinh ^{-1}(y) + \sinh ^{-1}(z))]) \\<br /> =\sinh [\sinh ^{-1} (x) + \sinh ^{-1} [\sinh (\sinh ^{-1}(y))\cosh (\sinh ^{-1} (z))+\cosh (\sinh ^{-1} (y))\sinh (\sinh ^{-1} (z))]]

Le's go into a different direction:

B:
x\ast (y\ast z) = \sinh (\sinh ^{-1} (x)+\sinh ^{-1}(y\ast z)) \\<br /> =\sinh (\sinh ^{-1}(x)+\sinh ^{-1} [\sinh (\sinh ^{-1}(y) + \sinh ^{-1}(z))]) \\<br /> =\sinh (\sinh ^{-1}(x)+ \sinh ^{-1}(y) + \sinh ^{-1}(z))

EDIT: Ah. too late.

 

[Dick]

whoops! lol didnt see you down here, was just trying to fix my tex :) *looking at second line*

Umm maybe I don't see what you mean, I can't simplify directly when there is a sum involved, right?

arcsinh(sinh( arcsinh(x)+arcsinh(y) ))=arcsinh(x)+arcsinh(y) is what I and ILS mean.

 

[ArcanaNoir]

Le's go into a different direction:

B:
x\ast (y\ast z) = \sinh (\sinh ^{-1} (x)+\sinh ^{-1}(y\ast z)) \\<br /> =\sinh (\sinh ^{-1}(x)+\sinh ^{-1} [\sinh (\sinh ^{-1}(y) + \sinh ^{-1}(z))]) \\<br /> =\sinh (\sinh ^{-1}(x)+ \sinh ^{-1}(y) + \sinh ^{-1}(z))

EDIT: Ah. too late.

how did you get from your second line to your third line?

 

[ArcanaNoir]

arcsinh(sinh( arcsinh(x)+arcsinh(y) ))=arcsinh(x)+arcsinh(y) is what I and ILS mean.

What identity are you using?

 

[Dick]

What identity are you using?

The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.

 

[ArcanaNoir]

The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.

but is it true that sinh(a+b)= sinh(a)+sinh(b)?

 

[Dick]

but is it true that sinh(a+b)= sinh(a)+sinh(b)?

Definitely not! That's not what I'm doing. Look at it again.

 

[ArcanaNoir]

The most basic one. sinh(arcsinh(A))=A. sinh and arcsinh are inverse functions. It doesn't matter that A is a sum.

arcsinh(sinh( arcsinh(x)+arcsinh(y) ))=arcsinh(x)+arcsinh(y) is what I and ILS mean.

Honestly I don't see how you are getting there using only sinh(arcsinh(A))=A. Surely you are using an additional identity.

 

[Dick]

Honestly I don't see how you are getting there using only sinh(arcsinh(A))=A. Surely you are using an additional identity.

A=arcsinh(x)+arcsinh(y). This might be a little hard to see because it's really simple and you are expecting something complicated.

 

[ArcanaNoir]

A=arcsinh(x)+arcsinh(y). This might be a little hard to see because it's really simple and you are expecting something complicated.

ohhhh. thank you :) I was trying to work from the inside out.

 

[ArcanaNoir]

Holy crap, here it goes! Maybe I can be done with this problem now!

A:
(x*y)*z = sinh[arcsinh(x)+arcsinh(y)]*z
=sinh[arcsinh(sinh[arcsinh(x)+arcsinh(y)])+arcsinh(z)]
=sinh[arcsinh(x)+arcsinh(y)+arcsinh(z)]

B:
x*(y*z) = sinh(arcsinh(x)+arcsinh[y*z])
=sinh(arcsinh(x)+arcsinh[sinh(arcsinh(y)+arcsinh(z))])
=sinh(arcsinh(x)+arcsinh(y)+arcsinh(z))

Yeah?

 

[Ray Vickson]

Okay, you guys have been wonderful, especially Dick and Micro. Although I'm beginning to suspect Micro is a genius, in which case he sneezes and answers pop out of him, so it's not like he has to try really hard. :P j/k I appreciate your tireless efforts Micro! (not kidding about thinking you're a genius...)
Anyway I'm trying to use Micro's hint but I feel like I'm slipping past the part where I'm supposed to rearrange the expressions to show they are equivalent. I will show my simplification for the expressions, maybe someone can point out where I was supposed to do something trig-y.

I have verified that \cosh (\sinh ^{-1}(x))=\sqrt{1+x^2} and that f(x,y)=\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))

A:
(x\ast y)\ast z = \sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))\ast z \\ <br /> = \sinh (\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]+\sinh ^{-1} (z)) \\<br /> =\sinh [\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]]\cosh (\sinh ^{-1} (z))+\\<br /> \cosh [\sinh ^{-1} [\sinh (\sinh ^{-1}(x)+\sinh ^{-1}(y))]]\sinh (\sinh ^{-1} (z)) \\ <br /> =\sinh [\sinh ^{-1} [\sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1} (y))+\cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1} (y))]]\cdot \sqrt{1+z^2} +\\<br /> \cosh [\sinh ^{-1} [ \sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1}(y))+\cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1} (y))]]\cdot z \\<br /> \sinh [\sinh ^{-1} [x\sqrt{1+y^2}+y\sqrt{1+x^2}]]\sqrt{1+z^2}+\cosh [\sinh ^{-1} [x\sqrt{1+y^2}+y\sqrt{1+x^2}]]z

B:
x\ast (y\ast z) = \sinh (\sinh ^{-1} (x)+\sinh ^{-1}(y\ast z)) \\<br /> =\sinh (\sinh ^{-1}(x)+\sinh ^{-1} [\sinh (\sinh ^{-1}(y) + \sinh ^{-1}(z))]) \\<br /> =\sinh [\sinh ^{-1} (x) + \sinh ^{-1} [\sinh (\sinh ^{-1}(y))\cosh (\sinh ^{-1} (z))+\\<br /> \cosh (\sinh ^{-1} (y))\sinh (\sinh ^{-1} (z))]] \\<br /> =\sinh [ \sinh ^{-1}(x)+\sinh ^{-1} [y\sqrt{1+z^2}+z\sqrt{1+y^2}]] \\<br /> =\sinh (\sinh ^{-1} (x))\cosh (\sinh ^{-1} [y\sqrt{1+z^2}+z\sqrt{1+y^2}])+\\<br /> \cosh (\sinh ^{-1} (x))\sinh (\sinh ^{-1}[y\sqrt{1+z^2}+z\sqrt{1+y^2}]) \\<br /> =x\sqrt{1+(y\sqrt{1+z^2}+z\sqrt{1+y^2})^2}+[y\sqrt{1+z^2}+z\sqrt{1+y^2}]\sqrt{1+x^2}

So it seems to me that this isn't going to end any better than when I didn't use hyp. trig, Hence why I think I'm missing the critical point.

No! From $f(u,v) = g\left( g^{-1}(u) + g^{-1}(v) \right)$, where $g(.) = \sinh(.)$,
it follows that
f(f(x,y),z) = g\left( g^{-1}(f(x,y)) + g^{-1}(z)\right) = g \left( g^{-1}(x) + g^{-1}(y) + g^{-1}(z) \right) .

 

[ArcanaNoir]

No! From $f(u,v) = g\left( g^{-1}(u) + g^{-1}(v) \right)$, where $g(.) = \sinh(.)$,
it follows that
f(f(x,y),z) = g\left( g^{-1}(f(x,y)) + g^{-1}(z)\right) = g \left( g^{-1}(x) + g^{-1}(y) + g^{-1}(z) \right) .

Yeah... I got that finally.

 

[ArcanaNoir]

Another huge thank you to everyone who helped with this problem, especially Ray, Dick, and Micro, who invested quite some time in helping me. I really really appreciate it, I couldn't have done it without all the outstanding help!